CHAPTER 5 Magnetism _ Electromagnetic Induction

CHAPTER 5

MAGNETISM &
ELECTROMAGNETIC INDUCTION

• 5.1 Magnetism
• 5.2 Magnetic field produced by current-carrying conductor.
• 5.3 Magnetic flux
• 5.4 Induced emf

Magnetism &
Electromagnetic

Induction

Magnetism Electromagnetic
Induction

Magnetic field Magnetic of Magnetic Flux Induce emf
sources current- ϕ = • Ԧ
carrying
Straight wire conductor
=
Circular Coil Solenoid Magnetic flux Faraday's Law
2 linkage Φ
= = Lenz's Law Formula
2 Φ = ϕ = −

in straight in coil
conductor

= = − ,


=

Magnetism &
Electromagnetic

Induction

Magnetism Electromagnetic
Induction

Magnetic field Magnetic of Magnetic Flux Induce emf
sources current- ϕ = • Ԧ
carrying
Straight wire conductor
=
Circular Coil Solenoid Magnetic flux Faraday's Law
2 linkage Φ
= = Lenz's Law Formula
2 Φ = ϕ = −

in straight in coil
conductor

= = − ,


=

5.1 Magnetism

a) Define magnetic field.
b) Identify magnetic field sources.

*Example:
i. Bar magnet and current-carrying conductor (straight wire and circular coil).
ii. Earth magnetic field.

Magnetic field Opposite
Poles
• Definition: attract

A region around a magnet or a current carrying Same
conductor where magnetic force is Poles
experienced. repel
• Magnetic field has two poles, called north (N)

and south (S) Figure 1. This magnetic poles
are always found in pairs whereas a single
magnetic pole has never been found.

• Opposite poles (N-S) attract each other
Figure 1 (a), (b).

• Like poles (N-N or S-S) repel each other
Figure 1 (c ), (d).

Figure 1

Magnetic Field Figure 2 (a): Direction Figure 2 (b): Direction
of magnetic field. of magnetic field.
• Is a vector quantity & its direction can be
determined by using a small compass X XXX Figure 3 (b): Magnetic
needle Figure 2 (a). field lines leave (out of)
X XXX the page perpendicularly
• Direction of the magnetic field
i. always tangent to the magnetic Figure 3 (a): Magnetic
field lines field lines enter (into) the
ii. exists the North pole Figure 2 (b). page perpendicularly
iii. enter the South pole Figure 2 (b).

• Magnetic field lines are used to
represent a magnetic field.

• Magnetic field can be represented by
crosses or by dotted circles as
shown in Figures 3(a) and 3(b).

Magnetic field sources (a) (b)

• Earth ( B = 5×10-5 T ) → natural permanent
magnet Figure 4(a).

• Bar magnet – permanent magnet with a typical
magnetic field Figure 4(b).

• Current carrying wire (conductor)
• Long straight wire Figure 4(c)
• Circular coil (loop) Figure 4 (d)
• Solenoid Figure 4(e)

(c) (d) (e)

Figure 4: magnetic field source

Magnetic field lines patterns
Earth

 The Earth’s magnetic field is like that of a 11.5
giant bar magnet as illustrated in Figure 5
with a pole near each geographic pole of the
 Earth. South
The magnetic poles are tilted away from the magnetic pole

rotational axis by an angle of 11.5.

 Since the north pole of a compass needle
(Figure 5) points toward the south magnetic
pole of the Earth, and since opposite attract,
it follows that

 the north geographical pole of the Earth is South North
actually near the south pole of the Earth’s geographical pole magnetic pole
magnetic field.

 Figure 5 also shows that the field lines are
essentially horizontal (parallel to the Earth’s
surface) near the equator but enter or leave
the Earth vertically near the poles.

Figure 5

Magnetic field lines patterns
Bar magnet

Figure 6

Magnetic field lines patterns
Bar magnet

Figure 7

EXERCISE 1:

Sketch the magnetic field lines pattern around the bar magnets for following cases.
a. b.

5.2 Magnetic field produced by current-carrying
conductor.

• a) Sketch magnetic field lines patterns of current carrying straight wire, circular
coil and solenoid.

• b) Calculate the magnetic field by using:

• i) for a long straight wire =

2

• ii) at the centre of a circular coil =

2

• iii) at the centre of solenoid =



Any CURRENT CARRYING conductor can produced magnetic field.

Magnetic field lines Figure 8 (a)
Figure 8 (b)
• A magnetic field can be represented by
a set of straight lines or curves
Figure 8 (a).

• Magnetic field lines exit from the
north pole of a magnet and enter at
the south pole Figure 8 (b).

• Magnetic field lines do not intersect
one another.

• Magnetic field also can be
represented by crosses or by dotted
circles as shown in Figure 3 (a) &(b).

Magnetic field lines

• A uniform field is represented by parallel A1 A2
lines of force, straight and equal space. This Figure 9: Uniform field
means that the number of lines passing
perpendicularly through unit area at all cross- A1 A2
sections in a magnetic field are the same as Figure 10: Non uniform field
shown in Figure 9 (A1 & A2.)

• Magnetic field is strong when the lines are
more and close each others Figure 10 A1,
less lines and far indicates weak magnetic field
Figure 10 A2.

Magnetic Field produced by current-
carrying conductor

• When a current flows in a conductor Figure 11
wire or coil, the magnetic field will be
produced.

• The direction of magnetic field
around the wire or coil can be
determined by using the right hand
grip rule as shown in Figure 11.

• Thumb – direction of current

• Other fingers – direction of
magnetic field (clockwise OR
anticlockwise)

Magnetic field of a long straight
conductor (wire) carrying current

• The magnetic field lines pattern around a straight conductor
carrying current is shown in Figures 12 (a) and 12 (b).

BI 
IB
 
IB B  OR XI

O
R
B
 IB
BI B

Current out of the X Current into the

page page

(a) (b)
Figure 12

• Consider a straight conductor (wire) r XPB
carrying a current I is placed in vacuum
as shown in Figure 13. into the page (paper)

• At a perpendicular distance r from a long I
straight wire carrying a current I, the
magnitude of magnetic field, B is given Figure 13
by:
From : B = o I
B = μ0I 2 r
2r ↑↑
; ↑  1
where μ0 : permeability of free space = 4 10−7 T m A−1 BI
r :distance of a point from a straight conductor (wire) B r↓

The magnetic field, B becomes stronger as the radial
distance r decreases, so the field lines are closer together
near the wire.

Example 1:
A long wire (X) carrying a current of 50 A is placed parallel to and 5.0
cm away from a similar wire (Y) carrying a current of 10 A.
a. Determine the magnitude and direction of the magnetic flux

density at a point midway between the wires :
i. when the current are in the same direction.
ii. when they are in opposite direction.
b. When the currents are in the same direction there is a point
somewhere between X and Y at which the magnetic flux density
is zero. How far from X is this point ?

(Given 0 = 4  10−7 H m−1)

Solution : I X = 50 A; d = 5.0 10−2 m; IY =10 A 
a. i. d BX

BX

rX A rY OR rX A rY
 IX  IY
BY
BY

IX IY rX = rY = d = 2.5 10−2 m
2

By using the equation of magnetic field at any point near the

straight wire, then at point A

Magnitude of BX : μ0 I X ( ( ) )BX
2πrX
BX = = 4π 10−7 50
2π 2.5 10−2

BX = 4.0 10−4 T

Direction : into the page OR upwards

Solution : IX = 50 A;d = 5.0 10−2 m; IY = 10 A
a. i. Magnitude of BY :
( ( ) )BY
BY = μ0 I Y = 4π 10−7 10
2πrY 2π 2.510−2

BY = 8.0 10−5 T

Direction : out of page OR downwards

Therefore the total magnetic flux density atpoint A is
BA = BX + BY

Note: BA = −BX + BY BA = −4.0 10−4 + 8.0 10−5
Sign convention of B: BA = −3.2 10−4 T

Out of the page  positive (+) Direction : into the page OR upwards

Into the page  negative (−)

Solution : IX = 50 A;d = 5.0 10−2 m; IY = 10 A 
a. ii. d
 BX  BX
BY
rX ABY rY OR rX A rY X

IX IY

IX IY

By using the equation of magnetic field at any point near the

straight wire, then at point A

( ( ) )Magnitude of BX :
BX = 4π 10−7 50 BX = 4.0 10−4 T
2π 2.5 10−2
Direction : into the page OR

upwards

Solution : IX = 50 A;d = 5.0 10−2 m; IY = 10 A

a. ii. Magnitude of BY :

( ( ) )BY
= 4π 10−7 10
2π 2.5 10−2

BY = 8.0 10−5 T

Direction : into the page OR upwards

Therefore the resultant magnetic flux density at point A is
BA = BX + BY

BA = −BX − BY BA = −4.0 10−4 − 8.0 10−5
BA = −4.810−4 T

Direction : into the page OR upwards

Solution : IX = 50 A;d = 5.0 10−2 m; IY = 10 A 
BX
b. d 
BX

rX C rY OR rX C rY
 IX  IY
BY
BY
IX IY rX = r
rY = d − r

Since the resultant magnetic flux density at point C is zero

thus   
BC = BX + BY

0 = −BX + BY BX = μ0 I X and BY = μ0 I Y
BX = BY where 2πrX 2πrY

Solution : IX = 50 A;d = 5.0 10−2 m; IY = 10 A

b. μ0 I X = μ0 I Y
2πrX 2πrY

IX = IY
r
(d − r)

10
5.0 10−2 − r
( )50 =

r

r = 4.2 10−2 m

EXERCISE 2:

Two long parallel wires carry currents of 8 A and 2 A. What is the
magnetic field and it direction at point A and B caused by wire 1 and
wire 2?

A

Magnetic field due to current flowing in a
circular coil

• The magnetic field lines pattern around a circular coil carrying
current is shown in Figures 15.

I View from above

N or N

S O ●I ×I
R
II S

Figure 15

• Figure 16 (a) show the Magnetic Field
magnetic field at the (North)
center of the coil.
I
• Used right hand rule to Curl Fingers – I
find direction of Thumb point – North Pole
magnetic field pattern
B Figure 16 (b) and (c)
(c).
Figure 16

• Consider a circular shaped R

conductor with radius R that O
carries a current I as shown in
Figure 17
Figure 17.
• The magnitude of magnetic field

intensity B at point O (centre of
the circular coil or loop) , is given
by

B = μ0 NI
2R

where μ0 : permeability of free space
R : radius of the circular coil
N : number of coils (loops)
I : current

Example 2:

A closely wound circular coil of diameter 10 cm has 500 turns and
carries a current of 2.5 A. Determine the magnitude of the magnetic

field at the centre of the coil. (Given 0 = 4  10−7 T m A−1)

Solution : Given R = 10 10−2 = 5.010−2 m; N = 500; I = 2.5 A
2

By applying the equation for magnitude of the magnetic field at the

centre of the circular coil, thus

( ( ) )B =
B = μ0 NI 4π 10−7 (500)2.5
2R
2 5.0 10−2

B = 1.57 10−2 T

EXERCISE 3:

A circular coil has 15 turns and a diameter of 45.0 cm. If the magnetic
field strength at the center of the coil is 8.0×10−4 T, find the current
flowing in the coil.

Magnetic field for current carrying
solenoid

• A solenoid is an electrical device in which a long wire has
been wound into a succession of closely spaced loops with
geometry of a helix Figure 18.

Figure 18: Coil

N OR S

I XI XI The magnetic field lines
I pattern around a solenoid
XI II
carrying current is shown
N in Figure 19.

I XI

S

I

Figure 19

The magnitude of magnetic field strength at The magnitude of magnetic field
the centre (mid-point/inside) of N turn strength at the end of N turn
solenoid is given by solenoid is given by

 = o NI  B = 1 onI
B
L B = on I 2

N Where,
n is the number of turns per unit length
and L = n

n is the number of turns per unit length
N = total number of turns
L = length of the solenoid

Applying the right hand grip S(a) N

rule to determ ine the I I
direction of B .
S
N
I
I IMPORTANT N(b)

Thumb – north pole I
Other fingers –
direction of current in Figure 20
solenoid.

Figure 21

Example 3:

A solenoid of length 1.5 m and 2.6 cm in diameter carries a
current of 18 A. The magnetic field inside the solenoid is 2.3
mT. Calculate the length of the wire forming the solenoid.

(Given 0 = 4  10−7 T m A−1)

Solution : 2.6 10−2
2
Given l = 1.5 m; r = = 1.310−2 m; Bi = 2.310−3 T; I = 18 A

By applying the equation of magnetic flux density inside the

solenoid, thus μ0 NI ( )2.310−3 = 4π 10−7 N(18)
l 1.5
Bi = N =153 turns

Solution :

Since the shaped for each coil in the solenoid is circle, then the

circumference for one turn is

( )circumference = 2πr circumference = 2π 1.310−2
circumference = 8.17 10−2 m

Therefore the length of the wire forming the solenoid is

L = N  (circumference)

( )L =153 8.1710−2

L =12.5 m

EXERCISE 4:

A solenoid of 50 turns is carrying a current of 10 mA. The magnetic
field strength at the center is 1.05×10−6 T. Calculate the length of the
solenoid.

SUMMARY CONCEPT

Current can produced magnetic field.

(1) Long straight wire B = o I
2 r

(2) Circular coil / loop B = oNI

2R

(3) Solenoid Bcentre = on I

and Bend = 1 onI

2

Direction B can be determined using right hand rule.

5.3 Magnetic flux

• a) Define and use magnetic flux, ϕ = • Ԧ
• b) Use magnetic flux linkage, Φ = ϕ

Magnetic flux ( ΦB )

Magnetic flux is a measure of the number of magnetic
field lines that cross a given area.


B

Figure 22

If a magnetic field of strength B crosses an area A at an
angle θ relative to the area’s normal figure 23, the
magnetic flux is :


 = B•A

 = B A cos



 : angle between Normal A and B

Figure 23

-- Unit SI for Φ : T m2 or Wb ( weber)
-- is a scalar quantity

Loop is perpendicular to the field Figure 24(i)
θ = 0 ° → Φ = BA (maximum)
Magnetic filed B crosses a surface area A at right angle.

i ii iii
Figure 24

Loop is parallel to the magnetic field Figure24(iii)
θ = 90 ° → Φ = 0
No field lines cross the surface

Magnetic flux linkage
If the coil is composed of N turns figure 25 all of the same area A,
thus the magnetic flux through N turns coil (magnetic flux linkage ) is :

 = N B A cos

where Figure 25
B : magnetic field strength ( unit : T )
A : area of the coil ( unit : m 2 )
θ : angle between B and normal of area A
N : numbers of turns for the coil

Example 4:

A single turn of circular coil with a diameter of 3.0 cm is placed in the
uniform magnetic field. The plane of the coil makes an angle 30 to the
direction of the magnetic field. If the magnetic flux through the area of
the coil is 1.20 mWb, calculate the magnitude of the magnetic field.

Solution : d = 3.0 10−2 m; = 1.20 10−3 Wb

B 30

30 Ɵ

30

The area of the coil is coil ( )A =  3.0 10−2 2
4
A = d 2 A = 7.07 10−4 m2

4

Solution : d = 3.0 10−2 m;  = 1.20 10−3 Wb

The angle between the direction of magnetic field, B and vector of area,
A is given by

Ɵ= 90∘ − 30∘ = 60∘

Therefore the magnitude of the magnetic field is
Φ = cos Ɵ

( )1.20 10−3 = B 7.07 10−4 cos60

B = 3.40 T

Example 5:

A 20 turn of rectangular coil of sides 10 cm  5.0 cm is placed between
north and south poles of a permanent magnet. Initially, the plane of the
coil is parallel to the magnetic field as shown in below.

N SR Q
II
SP

If the coil is turned by 90 about its rotation axis and the magnitude of
magnetic flux density is 1.5 T, Calculate the change in the magnetic flux
through the coil.

Solution : B =1.5 T

The area of the coil is

( )( )A = 1010−2 5.010−2 = 5.010−3 m2

Initially, A  From the figure,  =90 thus the initial magnetic flux
B
through the coil is Φi = cos Ɵ
= cos 9 0∘

Φi = 0

Finally,  From the figure,  =0 thus the final magnetic flux
B
 through the coil is Φf = cos Ɵ
A = 20 1.5 5.0 × 10−3 cos 0∘

Φf = 1.5 × 10−3 Wb

Therefore the change in magnetic flux through the coil is
Φ = Φf − Φi ΔΦ = 1.5 × 10−3 − 0
ΔΦ = 1.5 × 10−3 Wb